Understanding Energy Transfer and Calorimetry
Temperature is a measure of the average kinetic energy of particles in a substance.
Heat is the total energy transferred between substances due to a temperature difference.
\[q = mc\Delta T\]
q = heat energy (Joules, J)
m = mass of substance (grams, g)
c = specific heat capacity (J/g°C)
ΔT = change in temperature = \(T_{\text{final}} - T_{\text{initial}}\) (°C)
Specific heat is the amount of energy needed to raise 1 gram of a substance by 1°C.
| Substance | Specific Heat (J/g°C) |
|---|---|
| Water | 4.18 |
| Aluminum | 0.89 |
| Copper | 0.385 |
| Iron | 0.449 |
| Lead | 0.128 |
Problem: How much heat is needed to raise 250 g of water from 20°C to 80°C?
Given:
Solution:
\[q = mc\Delta T = (250)(4.18)(60) = 62,700 \text{ J} = 62.7 \text{ kJ}\]
Answer: 62,700 J or 62.7 kJ
Problem: A 100 g piece of aluminum at 25°C absorbs 5000 J of heat. What is its final temperature?
Given:
Solution:
\[q = mc\Delta T\]
\[5000 = (100)(0.89)(T_f - 25)\]
\[5000 = 89(T_f - 25)\]
\[56.18 = T_f - 25\]
\[T_f = 81.18°C \approx 81°C\]
Answer: 81°C
Thermal equilibrium occurs when two objects in contact reach the same temperature.
In an isolated system (no heat lost to surroundings):
\[\text{Heat lost by hot object} = \text{Heat gained by cold object}\]
\[q_{\text{hot}} = -q_{\text{cold}}\]
\[q_{\text{lost}} + q_{\text{gained}} = 0\]
Problem: A 50 g piece of copper at 95°C is dropped into 200 g of water at 20°C. What is the final temperature? (Assume no heat loss to surroundings)
Given:
Solution:
Heat lost by copper = Heat gained by water
\[-q_{\text{Cu}} = q_{\text{H₂O}}\]
\[-m_{\text{Cu}} c_{\text{Cu}} \Delta T_{\text{Cu}} = m_{\text{H₂O}} c_{\text{H₂O}} \Delta T_{\text{H₂O}}\]
\[-(50)(0.385)(T_f - 95) = (200)(4.18)(T_f - 20)\]
\[-(19.25)(T_f - 95) = (836)(T_f - 20)\]
\[-19.25T_f + 1828.75 = 836T_f - 16,720\]
\[1828.75 + 16,720 = 836T_f + 19.25T_f\]
\[18,548.75 = 855.25T_f\]
\[T_f = 21.7°C\]
Answer: 21.7°C
Notice: The final temperature is much closer to the water's initial temperature because water has a much larger mass AND higher specific heat!
A calorimeter is an insulated device used to measure heat transfer between substances.
Procedure:
Since the density of water is 1.0 g/mL:
\[\text{Mass of water (g)} = \text{Volume of water (mL)} \times 1.0 \text{ g/mL}\]
Example: 150 mL of water = 150 g of water
Problem: A 65.0 g metal sample is heated to 100°C and placed in 100.0 mL of water at 21.5°C. The final temperature is 26.8°C. What is the specific heat of the metal?
Given:
Solution:
Step 1: Calculate heat gained by water
\[q_{\text{water}} = m c \Delta T = (100.0)(4.18)(26.8 - 21.5)\]
\[q_{\text{water}} = (100.0)(4.18)(5.3) = 2215.4 \text{ J}\]
Step 2: Heat lost by metal = Heat gained by water
\[q_{\text{metal}} = -2215.4 \text{ J}\]
Step 3: Solve for specific heat of metal
\[q_{\text{metal}} = m c \Delta T\]
\[-2215.4 = (65.0)(c)(26.8 - 100)\]
\[-2215.4 = (65.0)(c)(-73.2)\]
\[-2215.4 = -4758c\]
\[c = \frac{2215.4}{4758} = 0.466 \text{ J/g°C}\]
Answer: 0.47 J/g°C (rounded to 2 sig figs)
Heat ALWAYS flows from hot to cold, never the reverse (without external energy input)
In an isolated system: Heat lost = Heat gained
When two objects reach the same temperature, no more net heat transfer occurs
Water has high specific heat (4.18 J/g°C), making it resistant to temperature change
Insulated device to measure heat transfer while minimizing energy loss to surroundings
Metal loses heat to air during transfer, affecting experimental accuracy